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S-110
A. What does the term Interference mean when applied to waves?
B. Describe what you think would happened when light interferes constructively.
C. Describe what would happen if light went through total destructive interference.
Objectives
I can explain Young’s double slit experiment.
I can calculate wavelength using interference patterns.
Interference and Diffraction
9.1 Superposition and Interference
9.1 Superposition and Interference
Review – Superposition
net displacement caused by a combination
of waves – the algebraic sum
Phase difference
180o – out of step by half a wavelength
Coherent – maintains a constant phase difference
Applet
9.2 Young’s Two-Slit Experiment
9.2 Young’s Two-Slit Experiment
1801 – Thomas Young
Shot monochromatic, coherent light
through two slits
The result was bands of color
called bright fringes
Applet
9.2 Young’s Two-Slit Experiment
What happened
Hugyen’s principle – every point on a wave front can be treated as the source of a new wave
When a wave is sent
through a barrier
two identical wave
fronts are created
9.2 Young’s Two-Slit Experiment
So when a wave is sent through double slits
The fringes are caused by
constructive interference – bright fringes
destructive interference – dark fringes
When we look at the pathway
of two rays
Ripple Tank
9.2 Young’s Two-Slit Experiment
For constructive interference, the path difference must be a multiple of So
m – any whole
number
dsin
md sin
9.2 Young’s Two-Slit Experiment
For dark fringes – the waves must be 180o out of phase, so
To measure , we need to go back to the experiment
An look at the triangle
One side is L, we will
call the other side
x
)(sin 21 md
x
L
9.2 Young’s Two-Slit Experiment
We can calculate the angle using the law of tangents
x
L
L
xtan
S-111
A light is shined through two slits that are 1.2 mm apart. If the third order minima is produced on a screen 1.5, how far from the central maxima will it be. Assume that the light used has a frequency of 7.2 x1014 hz.
S-112
A light is shined through two slits that are 0.50 mm apart. If the fourth order maxima is produced on a screen 0.80 m from the slits, how far from the central maxima will it be. Assume that the light used has a wavelength of 1.8 x10-7 m.
9.3 Interference in Reflected Waves
9.3 Interference in Reflected Waves
Waves that reflect of objects at different locations can interfere
Waves can go through a phase change due to reflection
9.3 Interference in Reflected Waves
When waves reflect at a boundary as they go
From higher n to lower n – no phase change
From lower n to higher n – 180o phase change
9.3 Interference in Reflected Waves
Lets look at what happens in an air wedge
There is no phase change
at the first boundary
There is a phase change
at the second boundary
The paths are essentially the same length in a real interference pattern (or they wouldn’t hit the same part of eye)
9.3 Interference in Reflected Waves
So the length of the path is
For constructive
interference
For destructive interference
tttl 221
21
mt
2
21
21
21 2
mt
9.3 Interference in Reflected Waves
Thin Films – soap bubbles or oil slicks
First ray (phase change)
Second ray (no
phase change)
Solving in terms of
211 l
tl 22
nvac
n
9.3 Interference in Reflected Waves
Combining we get
So for constructive interference
211 l tl 22 nvac
n
nn
tl
22
vacn
ntt
22
mnt
vac
212
9.3 Interference in Reflected Waves
And for destructive interference
mnt
vac
2
S-113
Light of wavelength 615 nm strikes the surface of an oil film (n=1.55) that is floating on water (n=1.33). The light strikes the surface of the oil at an angle of 22o. What is the minimum thickness of the oil that would produce a destructive interference pattern?
9.4 Diffraction
9.4 Diffraction
Waves diffract (bend) when they pass through barriers
Single Slit Diffraction – monochromatic light sent through a single slit will cause an interference pattern
The pattern occurs because of the diffraction of light around the edge of the slit
Diffraction
Applet
9.4 Diffraction
Similar to double slit geometry
If a is the width of
the slit, the
first minimum
would occur
In general
sina
ma sin
9.4 Diffraction
If the distance D is much greater than the slit distance y, then
we can use the
Approximation
Combining with
D
ytanD
ysin sin
D
y
ma sin
mD
ya
a
Dmy
9.5 Resolution
9.5 Resolution
Resolution – the ability to visually separate closed spaced objects
Depends on the aperture (size of lens)
In a slit the first dark fringe would
be
For a circular aperture of
diameter D produces a
central bright and a
dark fringe at an angle
a
sin
D
22.1sin
9.5 Resolution
Due to this interference, a point source of light will be viewed as a circular image
Rayleigh’s Criterion: If the first dark fringe of one circular diffraction patter passes through the center of a second pattern, the two sources responsible for the patterns will appear to be a single source.
Applet
9.5 Resolution
Examples: pixels on TV’s and Computer
9.6 Diffraction Gratings
9.6 Diffraction Grating
Diffraction Grating – a system with a large number of slits
3D glasses
Produced most often by
taking pictures of slits
and putting them on a slide
Produces sharp, widely spaced fringes
9.6 Diffraction Grating
Patterns are caused by multiple waves interfering
This example shows a
maxima produced by
10 slits
There would be other areas of constructive interference
9.6 Diffraction Grating
Because different colors of light have different wavelengths, and diffract differently
We can separate colors
Equations are the same
as double slits
Constructive
Destructive md sin
)(sin 21 md